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3 years ago

PLEASE HELP!! In triangle ABC, B= 120 degrees, a=10, c=18, find C

Mathematics

2 answers:

S_A_V [24]3 years ago

7 0

Add all of them then subtract from 180

sveta [45]3 years ago

7 0

Check the picture below. Notice that we can always solve for the cosine and simply use the sides alone to get an angle.

$\bf cos^{-1}\left(\cfrac{10^2+(\sqrt{604})^2-18^2}{2(10)(\sqrt{604})} \right)=\measuredangle C
\\\\\\
cos^{-1}\left(\cfrac{100+604-324}{20\sqrt{604}} \right)=\measuredangle C \\\\\\
cos^{-1}(0.77309903583)\approx \measuredangle C\implies 39.366998916677^o\approx \measuredangle C$

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"Solving Equations by Completing the Square: m^2+2m-48=-6

kvv77 [185]

$m^2+2m-48=-6 \\ \\ m^2+2m-48+6=0\\ \\m^2+2m-42=0\\ \\a=1, \ b= 2, \ c= -42 \\ \\\Delta = b^{2}-4ac = 2^{2}-4*1*(-42)= 4+168 =172 \\\\\sqrt{\Delta }=\sqrt{172} =\sqrt{4*43}=2\sqrt{43}\\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-2-2\sqrt{43}}{2}=\frac{2(-1-\sqrt{43})}{2}= -1-\sqrt{43} \\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} = \frac{-2+2\sqrt{43}}{2}=\frac{ 2(-1+\sqrt{43})}{2}= -1+\sqrt{43}$

5 0

3 years ago

Read 2 more answers

Pleeeeese help me as fast as you can!!!

Aleonysh [2.5K]

Answer:

A. $y=3x-10$

C. $y+6=3(x-15)$

Step-by-step explanation:

Given:

The given line is $6x+18y=5$

Express this in slope-intercept form $y=mx+b$ , where m is the slope and b is the y-intercept.

$6x+18y=5\\18y=-6x+5\\y=-\frac{6}{18}x+\frac{5}{18}\\y=-\frac{1}{3}x+\frac{5}{18}$

Therefore, the slope of the line is $m=-\frac{1}{3}$ .

Now, for perpendicular lines, the product of their slopes is equal to -1.

Let us find the slopes of each lines.

Option A:

$y=3x-10$

On comparing with the slope-intercept form, we get slope as $m_{A}=3$ .

Now, $m\times m_{A}=-\frac{1}{3}\times 3=-1$ . So, option A is perpendicular to the given line.

Option B:

For lines of the form $x=a$ , where, a is a constant, the slope is undefined. So, option B is incorrect.

Option C:

On comparing with the slope-point form, we get slope as $m_{C}=3$ .

Now, $m\times m_{C}=-\frac{1}{3}\times 3=-1$ . So, option C is perpendicular to the given line.

Option D:

$3x+9y=8\\9y=-3x+8\\y=-\frac{3}{9}x+\frac{8}{9}\\y=-\frac{1}{3}x+\frac{8}{9}$

On comparing with the slope-intercept form, we get slope as $m_{D}=-\frac{1}{3}$ .

Now, $m\times m_{D}=-\frac{1}{3}\times -\frac{1}{3}=\frac{1}{9}$ . So, option D is not perpendicular to the given line.

8 0

3 years ago

Please help :(<br> [math]

sergij07 [2.7K]

Hey!

------------------------------------------------

First Equation:

2x + 3 = -7

2x + 3 - 3 = -7 - 3 (Subtract 7 to both sides)

2x = -10

2x/2 = -10/2 (Divide 2 to both sides)

x = -5

------------------------------------------------

Second Equation:

4.5x - 7 = 20

4.5x + (-7) + 7 = 20 + 7 (Add 7 to both sides)

4.5x = 27

4.5x/4.5 = 27/4.5 (Divide 4.5 to both sides)

x = 6

------------------------------------------------

Third Equation:

-3x + 7 = 28

-3x + 7 - 7 = 28 - 7 (Subtract 7 to both sides)

-3x/-3 = 21/-3 (Divide -3 to both sides)

x = -7

------------------------------------------------

Answers:

2x + 3 = -7 → x = -5

4.5x - 7 = 20 → x = 6

-3x + 7 = 28 → x = -7

------------------------------------------------

Hope This Helped! Good Luck!

8 0

3 years ago

What is the equation of the line with (1,0) and (6,6)?

ch4aika [34]

Answer:

6x -5y = 6

Step-by-step explanation:

You can use the 2-point form of the equation of a line to find it.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (6-0)/(6-1)(x -1) +0

y = 6/5(x -1) . . . . . . point-slope form

y = (6/5)x -6/5 . . . . slope-intercept form

5y = 6x -6

6x -5y = 6 . . . . . . . standard form

7 0

3 years ago

If EF = 2x-7, FG = 4x-20, and EG = 21, find the values of x, EF, and FG.

Serggg [28]

Given -:

EF = 2X-7

FG = 4X-20

EG = 21

FIND OUT

EF =?

FG = ?

To proof

as given in the question

we have the value of EF = 2X-7 , FG = 4X-20 and EG = 21

Thus we have

EG = EF + FG

21 = 2X-7 + 4X- 20

21 = 6x - 27

48 = 6x

x = 8

Hence proved

Now put this value

$EF = 2\times 8 - 7\\EF = 9$

Now

$FG = 4\times 8 - 20\\FG = 12$

Hence proved

8 0

3 years ago